ch-19 compatibility mode
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CHAPTER - 19
THERMAL PROPERTIES
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Problems to be practiced :
Problems related to thermal stress & strain, heat flux andheat capacity
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ISSUES TO ADDRESS...
• How does a material respond to heat?
• How do we define and measure...
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-- eat capac ty--coefficient of thermal expansion
--thermal conductivity
--thermal shock resistance
• How do ceramics, metals, and polymers rank?
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General: The ability of a material to absorbheat.
HEAT CAPACITY
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increase the temperature of the material.
energy input (J/mol)
C =
dQ
dT
heat capacity
(J/mol-K)temperature change (K)
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Two ways to measure heat capacity:
-- Cp : Heat capacity at constant pressure.
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-- Cv : Heat capacity at constant volume.
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Heat capacity...
--
Heat Capacity Vs Temp
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--reaches a limiting value of
3R
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T K
Heat capacity, C v3R Cv= constant
gas constant = 8.31 J/mol-K
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Debye temperature(usually less than T room )
• Atomic view:--Energy is stored as atomic vibrations.
--As Temperature goes up, so does the avg. energy of
atomic vibration.
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HEAT CAPACITY: COMPARISON
• PolymersPolypropylenePolyethylenePolystyreneTeflon
cp (J/kg-K)
at room T
•
1925185011701050
cp: (J/kg-K)
Cp: (J/mol-K)
material
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• cp significantly larger for polymers.
Magnesia (MgO)Alumina (Al2O3)
Glass
• MetalsAluminum
SteelTungstenGold
900
486128138
i n
c r e a s i n g c
p 940775
840
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•
Materials change size when heating.T init
L init
THERMAL EXPANSION
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L final − L initial
L initial= α ( T final − T initial )
coefficient of
thermal expansion (1/K)
finalL final
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• Atomic view:
Mean bond length increases with Temp.
Bond energy
T 5 )
T 1 )
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Bond length (r)
i n c r e
a s i n g T
T1
r
(
r
(
T5bond energy vs bond lengthcurve is “asymmetric”
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• PolymersPolypropylenePolyethylenePolystyreneTeflon
145-180106-198
90-150126-216
α (10-6/K)
at room T
• Metals
Material
THERMAL EXPANSION:COMPARISON
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• CeramicsMagnesia (MgO)Alumina (Al 2O3)
Soda-lime glassSilica (cryst. SiO 2)
13.5
7.6 9 0.4
SteelTungstenGold
.
124.5
14.2
i n c r e a s i n g
• Q: Why does α generally decrease with increasing bond
energy?
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• General: The ability of a material to transfer heat.
• Quantitative:
q = −k dT
dx
temperature
gradient
thermal conductivity (J/m-K-s)
heat flux
(J/m2-s)
THERMAL CONDUCTIVITY
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• Atomic view: Atomic vibrations in hotter region carry
energy (vibrations) to cooler regions.
K = Kl+K
e
T2 > T1 T1
x1 x2heat flux
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k (W/m-K)
• Metals
AluminumSteelTungstenGold
24752
178315
Energy Transfer
By vibration of atoms andmotion of electrons
Material
THERMAL CONDUCTIVITY:
COMPARISON
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• PolymersPolypropylenePolyethylenePolystyreneTeflon
0.120.46-0.500.130.25
•Magnesia (MgO)Alumina (Al2O3)
Soda-lime glassSilica (cryst. SiO 2)
3839
1.71.4
i n
c r e a s i n g
By vibration/rotation of chainmolecules
By vibration of atoms
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• Occurs due to:
THERMAL STRESS
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----mismatch in thermal expansion.
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• Example Problem 19.1, p. 666, Callister 6e.--A brass rod is stress-free at room temperature (20C).
--It is heated up, but prevented from lengthening.
--At what T does the stress reach -172MPa?
Troom
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∆L
Lroom= εthermal = α(T − Troom )
LroomT
∆L
compressive σ keeps ∆L = 0 σ = E(−ε thermal ) = −Eα(T − Troom )
100GPa 20 x 10-6 /C
20CAnswer: 106C-172MPa
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• Occurs due to: uneven heating/cooling.
• Ex: Assume top thin layer is rapidly cooled from T1 to T2:
rapid quench
THERMAL SHOCK RESISTANCE
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doesn’t want to contract
tries to contract during cooling T2
T1
Tension develops at surface
σ = − E α ( T 1 − T 2 )
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• Reducing the thermal gradient by
– changing its temperature more slowly
– increasing the material's thermalconductivity
'
Thermal shock can be prevented by:
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thermal expansion
• Increasing its strength
• Increasing its toughness
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Temperature difference that
can be produced by cooling:
(T1 − T2 )fracture =
σ f
Eα (T1 − T2 ) =
quench rate
k
Critical temperature
difference
for fracture (set σ = σ
f
)
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• Result:
• Large thermal shock resistance when is large.
σfk
Eα
(quench rate ) for fracture ∝σ f k
Eα
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• Application:
Space Shuttle Orbiter
THERMAL PROTECTION SYSTEM
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Re-entry TDistribution
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reinf C-C(1650°C)
silica tiles(400-1260°C)
nylon felt, silicon rubber
coating (400°C)
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• Silica tiles (400-1260C):--large scale application
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--
microstructure:
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100 µ m
~90% porosity Si fibers bonded to one another during
heat treatment.
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• A material responds to heat by:
--increased vibrational energy
--redistribution of this energy to
achieve thermal e uilibrium.
SUMMARY
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Heat capacity:
--energy required to increase a unit
mass by a unit Temp.
--polymers have the largest values.
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• Coefficient of thermal expansion:--the stress-free strain induced by
heating by a unit Temp.--polymers have the largest
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.• Thermal conductivity:
--the ability of a material totransfer heat.
--metals have the largest values.
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Thermal shock resistance:
--the ability of a material to be
rapidly cooled and not crack.Maximize
σfk
Eα
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